Probability that a < b^2?

Let's choose numbers "a" and "b" randomly and uniformly from the entire set of real numbers. What is the probability that a < b^2?
Update: Duke:
If P(a < b^2) --> 1
then that would mean
P(a > b^2) --> 0
right?
Update 2: Interesting discussion so far, I just find it so counter intuitive that P(a>b^2) could actually be zero.
Update 3: Well would you believe it? I ran a simple program in matlab which generated a random gaussian variable N(0, σ) for a and b.
My approximate results were
p = P(a < b^2) ~= 0.7, when σ=1
p~= 0.9, when σ=10
p~=0.97, when σ=100
p~=0.999, when σ=1000
Interesting indeed.
Update 4: scythian: "almost all the time" and "all the time" are two different things. P(a > b^2) = 0 means it cannot happen. But we know that whatever "b" is, we can always find a value of "a" greater than b^2. That's what I meant by 0 being unintuitive. Nevertheless, it does... show more scythian: "almost all the time" and "all the time" are two different things. P(a > b^2) = 0 means it cannot happen. But we know that whatever "b" is, we can always find a value of "a" greater than b^2. That's what I meant by 0 being unintuitive. Nevertheless, it does appear that more than one method is leading to zero as the answer.
Update 5: OK Jered, but my point still stands. If you pick a random real from 0 to 10, the prob of an integer (or rational number) is zero BECAUSE these are point events which have no area under the f(x) curve. This is NOT the case with a > b^2. For every b, there is a range of a that satisfies the inequality. In fact... show more OK Jered, but my point still stands. If you pick a random real from 0 to 10, the prob of an integer (or rational number) is zero BECAUSE these are point events which have no area under the f(x) curve.

This is NOT the case with a > b^2. For every b, there is a range of a that satisfies the inequality. In fact for every range of b, there is a range of a satisfying it. So it's not a point. It's a definite area with a non zero probability. However I agree that as the domain increases, the probability tends to zero.
Update 6: The same ideas hold for scythian's arctan(a/b) sector. The ratio of the area tends to zero, but the area itself is never really zero. That's what I find un-intuitive. I guess un-intuitive is a subjective word.
Update 7: But I guess that more or less clears up my doubts. In the case of picking a random real number between 0 and 10, P(integer) = 0 even though it's possible to pick an integer. Here P(a>b^2) --> 0 although it's still possible for it to happen.
Update 8: Now I have the unenviable task of picking a best answer.
Update 9: Merlyn: The consensus so far, as well as from Remo's question, is that it all depends on how you choose the points on the infinite plane. My feelings are that the answer should not be different if the method is different, otherwise we have a paradox (call it by some other name if you wish). Some of the... show more Merlyn: The consensus so far, as well as from Remo's question, is that it all depends on how you choose the points on the infinite plane. My feelings are that the answer should not be different if the method is different, otherwise we have a paradox (call it by some other name if you wish).

Some of the answerer's have chosen a,b in the domain (-N, N) and taken the limit as N --> ∞. They got 1. Now the other method is using a known infinite distribution and taking σ --> ∞. Again the result is 1.

Why is it a good idea? If nothing else, it confirms the answer using a different method.
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